iBoost alternatives? - Solar Photovoltaic (PV)

RE: iBoost alternatives?

Marzipan71 — Wed, 19 Apr 2023 09:19:41 +0000

@editor I watched the video - looks a great piece of kit. We have the same size PV array as you (6.1kWp) I think. Our average daily solar sent to grid in the winter months (Nov to Feb) is less than 4kWh so it might not be that useful for us in winter (on average, obviously there will be sunny days when it is) but the rest of the year when we are on average sending 8kWh a day to the grid I think it will be a great addition. Question I had from the video - sorry if I missed the information - how many kW is the immersion element in your main tank, and what is the volume? I will be limited to a 2kW immersion (I don't have one as yet but for my Daikin kit, its 2kW) in a 500L tank (thermal store). Thanks for the review!

RE: iBoost alternatives?

Mars — Tue, 11 Apr 2023 15:07:28 +0000

Our Immersun PowerDivert has now been running for a month, and we're incredibly impressed with the quality of this British made power diverter. Watch our review here: https://youtu.be/KuqiU97f6Hw

Spoiler alert: this product is far superior to the Solar iBoost.

RE: iBoost alternatives?

Derek M — Tue, 21 Mar 2023 10:09:20 +0000

We’re reviewing the Immersun, and I came across this formula that I used to calculate the amount of energy required to heat 300 liters of water from 25°C to 44°C:

Q = m * c * ΔT

Q is the amount of heat energy required, measured in joules (J)
m is the mass of water being heated, measured in kilograms (kg)
c is the specific heat capacity of water, which is 4.184 J/g°C
ΔT is the change in temperature, measured in Celsius (°C) - our old heat pump friend.

First, we convert the volume of water from liters to kilograms, by multiplying it by the density of water, which is approximately 1 kg/liter:

m = 300 kg

Then, we can calculate the change in temperature:

ΔT = (44°C - 25°C) = 19°C

Now, we substitute these values into the formula:

Q = m * c * ΔT Q = 300 kg * 4.184 J/g°C * 19°C Q = 23,918.4 kJ

To convert kilojoules (kJ) to kWh, we divide by 3,600 (the number of seconds in an hour) and multiply by the conversion factor of 0.2778:

23,918.4 kJ ÷ 3,600 s/hour x 0.2778 kWh/kJ ≈ 1.98 kWh

So it would take approximately 1.98 kWh of electricity to heat 300 liters of water from 25°C to 44°C.

Sound right?

I get the electrical energy required to be approximately 6.6 kWh.

Where does the x 0.2778 come from?

I also get 23848.8 kJ rather than 23918.4 kJ

RE: iBoost alternatives?

Fazel — Tue, 21 Mar 2023 09:26:27 +0000

@editor

When in doubt, google is my friend.

3413

RE: iBoost alternatives?

Mars — Tue, 21 Mar 2023 08:47:42 +0000

We’re reviewing the Immersun, and I came across this formula that I used to calculate the amount of energy required to heat 300 liters of water from 25°C to 44°C:

Q = m * c * ΔT

Q is the amount of heat energy required, measured in joules (J)
m is the mass of water being heated, measured in kilograms (kg)
c is the specific heat capacity of water, which is 4.184 J/g°C
ΔT is the change in temperature, measured in Celsius (°C) - our old heat pump friend.

First, we convert the volume of water from liters to kilograms, by multiplying it by the density of water, which is approximately 1 kg/liter:

m = 300 kg

Then, we can calculate the change in temperature:

ΔT = (44°C - 25°C) = 19°C

Now, we substitute these values into the formula:

Q = m * c * ΔT Q = 300 kg * 4.184 J/g°C * 19°C Q = 23,918.4 kJ

To convert kilojoules (kJ) to kWh, we divide by 3,600 (the number of seconds in an hour) and multiply by the conversion factor of 0.2778:

23,918.4 kJ ÷ 3,600 s/hour x 0.2778 kWh/kJ ≈ 1.98 kWh

So it would take approximately 1.98 kWh of electricity to heat 300 liters of water from 25°C to 44°C.

Sound right?

RE: iBoost alternatives?

Derek M — Sun, 19 Mar 2023 23:25:34 +0000

@derek-m, how can I calculate how much water I’m heating per degree centigrade based on diverting 1kWh of solar generation?

Put another way, if I have a 300 litre water cylinder, how much power would I need to divert to raise the temperature by 1C?

It takes approximately 1.16W to heat 1 litre of water by 1C, so 300 litres will take 348W.

RE: iBoost alternatives?

Fazel — Sun, 19 Mar 2023 22:10:04 +0000

@editor calculator number 2 says 350w

https://bloglocation.com/art/water-heating-calculator-for-time-energy-power

RE: iBoost alternatives?

Mars — Sun, 19 Mar 2023 21:45:17 +0000

@derek-m, how can I calculate how much water I’m heating per degree centigrade based on diverting 1kWh of solar generation?

Put another way, if I have a 300 litre water cylinder, how much power would I need to divert to raise the temperature by 1C?

RE: iBoost alternatives?

Mars — Sun, 12 Mar 2023 21:19:02 +0000

@derek-m, we've switched to an Immersun this week – it's a terrific piece of kit. Well designed. Easy to install. And works ridiculously well.

RE: iBoost alternatives?

Jez — Sun, 12 Mar 2023 14:28:20 +0000

@iancalderbank hi

I'm interested in your unused iboost.

We got a givenergy inverter and battery will be fitted soon.

We have Bosch 8000 and an electric immersion.

Regards Jez

Contact me via email.

Jez_dean@hotmail.co.uk